Question

Solve and get 100 points.
Solve 8 = 2^2x − 1.

Answer 1

x = 2

Explanation:

You want to know the solution to the equation 8 = 2^(2x-1).

Solution

We can write both sides in terms of powers of 2, then equate the exponents:

2^3 = 2^(2x -1)

3 = 2x -1 . . . . . . . equate the exponents

4 = 2x . . . . . . . add 1

2 = x . . . . . . . divide by 2

The value of x is 2.

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Additional comment

The equation given in the problem statement is a linear equation equivalent to 8 = 4x -1. This has the solution x = 9/4 = 2.25.

We have presumed you are interested in an exponential equation. When the equation is 8 = 2^(2x) -1, the solution is x = log(3)/log(2) ≈ 1.58496, an irrational number. This suggests you want the solution to the exponential equation we showed above.

Parentheses are helpful when there is an exponent that contains math operations.

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Answer 2

If

`\sf 8 = 2^2 x - 1`

`\sf x = (9)/(4) \textsf{ or } 2 (1)/(4) \textsf{ or } 2.25`

If

`\sf 8 = 2^(2 x )- 1`

`\sf x \approx 1. 59`

Explanation:

If the question is:

`\sf 8 = 2^2 x - 1`

Add 1 on both sides:

`\sf 8 + 1 = 2^2 x - 1 + 1`

`\sf 9 = 2^2 x`

Since 2² = 4.

So,

`\sf 9 = 4 x`

Divide both sides by 4.

`\sf (9)/(4) = (4 x)/(4)`

`\sf x = (9)/(4) \textsf{ or } 2 (1)/(4) \textsf{ or } 2.25`

If the question is:

`\sf 8 = 2^(2 x) - 1`

Add 1 on both sides:

`\sf 8 + 1 = 2^(2 x )- 1 + 1`

`\sf 9 = 2^(2 x)`

Taking the logarithm of both sides. This gives us:

`\sf log_2(9) = \log_2(2^(2 x))`

Using the property of logarithms that
`\sf \log_a(b^c) = c \log_a(b)`, we can simplify the right-hand side of the equation:

`\sf \log_2(9) = 2 x \log_2(2)`

Since
`\sf \log_2(2) = 1`, we can further simplify the equation:

`\sf \log_2(9) = 2 x`

Finally, we can divide both sides of the equation by 2 to solve for x:

`\sf x = (\log_2(9))/(2)`

Using a calculator, we can find that
`\sf \log_2(9) \approx 3.169925`, so the solution to the equation is:

`\sf x \approx (3.169925)/(2) \\\\ \approx 1.5849625 \\\\ \approx 1. 59 \textsf{ (in 2 d.p.)}`

Therefore, the solution to the equation is x = 1.59 approximately.